Question

Find the order of the given element.

(c)$\mathbf{\left(}\begin{array}{cc}\mathbf{0}& \mathbf{-}\mathbf{1}\\ \mathbf{1}& \mathbf{1}\end{array}\mathbf{\right)}$ in$\mathit{G}\mathit{L}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{ℝ}\mathbf{)}$

Short answer

The order of$\left(\begin{array}{cc}0& -1\\ 1& 1\end{array}\right)$ in $GL(2,ℝ)$is 6.

Step-by-step solution

Step 1:GL(2,ℝ)GL(2,ℝ)

The group$\mathit{G}\mathit{L}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{ℝ}\mathbf{)}$ is a group of all $\mathbf{2}\mathbf{\times }\mathbf{2}$invertible matrices over the field of real numbers$ℝ$

Order of an element

Order of an element$\mathit{a}\mathbf{\in }\mathit{G}$ is defined as a positive integer n such that${\mathit{a}}^{\mathbf{n}}\mathbf{=}\mathit{e}$, where$\mathit{e}$ is the identity element in $\mathit{G}$.

Finding order of the given element

The given element of $GL(2,ℝ)$is .$\left(\begin{array}{cc}0& -1\\ 1& 1\end{array}\right)$ Now,

$\begin{array}{l}{\left(\begin{array}{cc}0& -1\\ 1& 1\end{array}\right)}^2=\left(\begin{array}{cc}0& -1\\ 1& 1\end{array}\right)\left(\begin{array}{cc}0& -1\\ 1& 1\end{array}\right)=\left(\begin{array}{cc}-1& -1\\ 1& 0\end{array}\right),\\ {\left(\begin{array}{cc}0& -1\\ 1& 1\end{array}\right)}^3=\left(\begin{array}{cc}-1& -1\\ 1& 0\end{array}\right)\left(\begin{array}{cc}0& -1\\ 1& 1\end{array}\right)=\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right),\\ {\left(\begin{array}{cc}0& -1\\ 1& 1\end{array}\right)}^4=\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{cc}0& -1\\ 1& 1\end{array}\right)=\left(\begin{array}{cc}0& 1\\ -1& -1\end{array}\right),\\ {\left(\begin{array}{cc}0& -1\\ 1& 1\end{array}\right)}^5=\left(\begin{array}{cc}0& 1\\ -1& -1\end{array}\right)\left(\begin{array}{cc}0& -1\\ 1& 1\end{array}\right)=\left(\begin{array}{cc}1& 1\\ -1& 0\end{array}\right),\\ {\left(\begin{array}{cc}0& -1\\ 1& 1\end{array}\right)}^6=\left(\begin{array}{cc}1& 1\\ -1& 0\end{array}\right)\left(\begin{array}{cc}0& -1\\ 1& 1\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right).\end{array}$

Therefore, the order of$\left(\begin{array}{cc}0& -1\\ 1& 1\end{array}\right)$ in $GL(2,ℝ)$is 6.